Mathematics • Year 9 • Unit 3 • Lesson 2

Hypotenuses in the Real World

Apply $c = \sqrt{a^2 + b^2}$ to everyday situations: ladders, TVs, pizza boxes, ramps and football pitches. Set up each problem by sketching, labelling the two legs, and substituting carefully into Pythagoras' theorem. Then explain in your own words why $\sqrt{a^2 + b^2}$ is not the same as $a + b$.

Apply · Real-World Maths

1. Word problems

Each problem has two perpendicular distances — the LEGS — and asks you to find the slanted distance (the HYPOTENUSE). Always sketch, label, substitute into $c^2 = a^2 + b^2$, then square-root and round to 2 d.p. unless told otherwise. Show your working — a single final answer with no working only earns half marks.

1.1 — Ladder against a wall. A painter's ladder is set up with its base 1.8 m from the wall and its top reaching 3.2 m up the wall.

(a) Sketch the situation and label the two legs.
(b) Find the length of the ladder to 2 d.p. 3 marks

Stuck? The ladder is the hypotenuse (it's slanted and longer than the leg lengths). Substitute legs 1.8 and 3.2 into $c^2 = a^2 + b^2$.

1.2 — TV diagonal. A rectangular TV screen is 56.7 inches wide and 31.9 inches tall. Manufacturers always quote the diagonal in inches.

(a) Find the diagonal of the screen to 2 d.p.
(b) Which whole-number TV size (the nearest inch) would this be sold as? 3 marks

Stuck? Use $c = \sqrt{56.7^2 + 31.9^2}$ — this is exactly the worked example in the lesson.

1.3 — Pizza box. Maya orders a square pizza in a box with side length 30 cm. The pizza shop wraps a flat cardboard label across the diagonal of the box.

(a) Find the length of the diagonal label to 2 d.p.
(b) Explain why the answer is greater than 30 cm even though both sides of the box are 30 cm. 3 marks

Stuck on (b)? The diagonal is the hypotenuse — and the hypotenuse is always the longest side of a right triangle, even when the two legs are equal.

1.4 — Skate ramp. A skate ramp has a horizontal base of 4 m and a vertical height at one end of 1.5 m. The slope of the ramp itself runs from the top of the vertical edge down to the far end of the base.

(a) Sketch the ramp (it forms a right triangle).
(b) Find the length of the slope to 2 d.p. 3 marks

Stuck? The slope of the ramp is the slanted side — that's the hypotenuse. The two legs are 4 m (horizontal) and 1.5 m (vertical).

1.5 — Football pitch diagonal. A school football pitch is 100 m long and 64 m wide. The PE teacher asks two students to run along the diagonal to test their fitness.

(a) Find the length of the diagonal run to 2 d.p.
(b) How much further is the diagonal run than the long side of the pitch (100 m)? 3 marks

Stuck? Diagonal of a rectangle = hypotenuse of a right triangle with legs equal to length and width. Subtract 100 m for part (b).

2. Explain your thinking

This question is about communication, not just answers. Use full sentences. 4 marks

2.1 A classmate writes "$\sqrt{9 + 16} = \sqrt{9} + \sqrt{16} = 3 + 4 = 7$". They use this on Pythagoras problems and keep getting wrong answers.

In your own words, explain (i) what the correct value of $\sqrt{9 + 16}$ is and how to get it, (ii) which rule from Lesson 2 they have broken, and (iii) what their wrong approach would give for the hypotenuse when the legs are 3 and 4 (and why we know this is wrong).

Stuck? Revisit lesson § "Spot the Trap" — the square root acts on the WHOLE sum, not term-by-term. And the 3-4-5 triple tells us the correct hypotenuse for legs 3 and 4.

How did this worksheet feel?

Got it Partly Lost

What I'll revisit before next class:

Answers — Do not peek before attempting

1.1 — Ladder against a wall

(a) Sketch: right triangle with horizontal leg 1.8 m (base distance) and vertical leg 3.2 m (height up the wall). The ladder is the slanted hypotenuse.
(b) $c^2 = 1.8^2 + 3.2^2 = 3.24 + 10.24 = 13.48$, so $c = \sqrt{13.48} \approx \mathbf{3.67}$ m. The ladder is about 3.67 m long.

1.2 — TV diagonal

(a) $c^2 = 56.7^2 + 31.9^2 = 3214.89 + 1017.61 = 4232.5$, so $c = \sqrt{4232.5} \approx \mathbf{65.06}$ inches.
(b) Nearest inch is 65, so it would be sold as a $\mathbf{65\text{-}inch}$ TV. (Manufacturers always round to a whole inch.)

1.3 — Pizza box diagonal

(a) $c^2 = 30^2 + 30^2 = 900 + 900 = 1800$, so $c = \sqrt{1800} \approx \mathbf{42.43}$ cm.
(b) The diagonal is the hypotenuse of a right triangle with two 30 cm legs. By Pythagoras' theorem the hypotenuse is always longer than either leg, because we add two positive numbers and then take the square root of a bigger total. (For a square specifically, the diagonal is always $\sqrt{2} \approx 1.41$ times the side length — so 30 cm sides give a 42.43 cm diagonal.)

1.4 — Skate ramp

(a) Sketch: right triangle with horizontal leg 4 m and vertical leg 1.5 m. The slope is the hypotenuse.
(b) $c^2 = 4^2 + 1.5^2 = 16 + 2.25 = 18.25$, so $c = \sqrt{18.25} \approx \mathbf{4.27}$ m. The slope is about 4.27 m long.
Real-world note: this is why measuring the "horizontal run" of a ramp underestimates how much wood you need — the slope is longer.

1.5 — Football pitch diagonal

(a) $c^2 = 100^2 + 64^2 = 10000 + 4096 = 14096$, so $c = \sqrt{14096} \approx \mathbf{118.73}$ m.
(b) Diagonal $-$ long side $= 118.73 - 100 = \mathbf{18.73}$ m further. (So the diagonal run is about 19 m further than just running the length of the pitch.)

2.1 — Explain your thinking (sample response)

(i) The correct value is $\sqrt{9 + 16} = \sqrt{25} = \mathbf{5}$. You must add INSIDE the square root first, then square-root the total — not square-root each piece separately. (ii) The classmate has broken the rule that the square root acts on the whole sum, not term-by-term (this is exactly the trap flagged in the lesson). (iii) With their wrong method applied to legs 3 and 4, they would get a "hypotenuse" of $\sqrt{3^2} + \sqrt{4^2} = 3 + 4 = 7$. We know this is wrong because the 3-4-5 Pythagorean triple from Lesson 1 tells us the correct hypotenuse is 5, not 7. Their answer would also disagree with measuring the side directly — a clear contradiction with reality.

Marking: 1 for the correct value of $\sqrt{9+16}=5$; 1 for naming the rule ("$\sqrt{}$ acts on the whole sum"); 1 for the wrong-method result of 7; 1 for explaining why this is wrong (3-4-5 triple gives 5, not 7).